An object of height 2 cm is placed at a distance 20 cm in front of a concave mirror of focal length 12 cm. Find the position, size and nature of the image.

Given,

Object height (O) = 2 cm

Focal length (f) = 12 cm (negative)

Object distance (u) = 20 cm (negative)

Mirror formula:

$\dfrac{1}{u} + \dfrac{1}{v} = \dfrac{1}{f}$

Substituting the values in the formula above we get,

$- \dfrac{1}{20} + \dfrac{1}{v} = - \dfrac{1}{12} \\[0.5em] \dfrac{1}{v} = - \dfrac{1}{12} + \dfrac{1}{20} \\[0.5em] \dfrac{1}{v} = \dfrac{ - 5 + 3 }{60} \\[0.5em] \dfrac{1}{v} = - \dfrac{ 2}{60} \\[0.5em] \dfrac{1}{v} = - \dfrac{1}{30} \\[0.5em] v = - 30 \text{ cm} \\[0.5em]$

Hence, the image is formed at a distance of 30 cm in front of the mirror.

$\text{Magnification (m)} = \dfrac{\text{Length of image (I)}}{\text{Length of object (O)}} \\[0.5em] = \dfrac{\text{Distance of image (v)}}{\text{Distance of object (u)}} \\[1em] \Rightarrow \dfrac{I}{2} = - \dfrac{30}{20} \\[0.5em] \Rightarrow I = -\dfrac{30 \times 2}{20} \\[0.5em] \Rightarrow I = - 3 \text { cm}\\[0.5em]$

Hence, length of image = 3 cm

Image will be real, inverted and magnified.